Expand ( x ^ 2 + 3 x ) ^ 4 , x 0

Question

Expand $\left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 } , x \neq 0$

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Answer

By using binomial theorem, we have
$(a + b)^n =\ ^nC_0a^n +\ ^nC_1a^{n-1} b +\ ^nC_2a^{n-2} b^2+ ... +\ ^nC_ra^{n-r}b^r + ... +\ ^nC_nb^n$
$\therefore \left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 } =\ ^4C_0(x^2)^4 +\ ^4C_1(x^2)^{4-1}\left( \frac { 3 } { x } \right) ^ { 1 } +\ ^4C_2(x^2)^{4-2} \left( \frac { 3 } { x } \right) ^ { 2 } +\ ^4C_3(x^2)^{4-3} \left( \frac { 3 } { x } \right) ^ { 3 }+ 4C_4(x^2)^{4-4} \left( \frac { 3 } { x } \right) ^ { 4 }$
$= 1.x^8 + 4 (x^2)^3\left( \frac { 3 } { x } \right) + 6 (x^2)^2 \left( \frac { 3 } { x } \right) ^ { 2 } + 4 (x^2) \left( \frac { 3 } { x } \right) ^ { 3 } + 1. (x^2)^0\left( \frac { 3 } { x } \right) ^ { 4 }$
$[\because\ ^4C_0 = 1,\ ^4C_1 = 4,\ ^4C_2 = 6,\ ^4C_3 = 4$ and $^4C_{4}= 1]$
$= x ^ { 8 } + 4 \cdot x ^ { 6 } \cdot \frac { 3 } { x } + 6 x ^ { 4 } \cdot \frac { 9 } { x ^ { 2 } } + 4 x ^ { 2 } \cdot \frac { 27 } { x ^ { 3 } } + 1 \cdot 1 \cdot \frac { 81 } { x ^ { 4 } }$
$= x^8 + 12x^5 + 54x^2 + \frac { 108 } { x } + \frac { 81 } { x ^ { 4 } }$

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